O.E.3: Identificar los desafíos que tiene Capittana con respecto a su competencia y consumidores

X

-ray diffraction is a powerful tool used to measure crystallinity and other lattice-dependent variables. X-ray diffraction also helps clarify the physical signifi cance of planes and Miller indices. The principle of X-ray diffraction has developed from the study of optics. Electromagnetic radiation (including X-rays and visible light) travels in waves. Each type of

2.7 | X-Ray Diffraction 45

| Interplanar Spacing | The distance between repeated planes in a lattice.

u u

(a) (b)

Figure 2-10 Interference Patterns for X-rays: (a) Constructive Interference;

(b) Destructive Interference

electromagnetic wave has a characteristic wavelength 1l2. The wavelengths in the X-ray range are roughly the same size as most interatomic distances. When a wave strikes a solid object (e.g., an atomic nucleus), it bounces off the object with an angle of refl ection equal to the angle of incidence. Figure 2-9 shows an X-ray beam striking atoms in a lattice.

Diffraction describes the interaction of waves. Figure 2-10 shows how two waves in phase add through constructive interference and two waves out of phase cancel through destructive interference.

In an X-ray diffraction machine like the one shown schematically in Figure 2-11, a source shoots X-rays into a sample and a detector collects the diffracted beams. The source and detector move together through different angles but always maintain the same “angle of incidence equals the angle of refl ection” relationship to each other. Because many different atoms are pre-sent in the lattice, most waves cancel. Net constructive interference results only whenBraggs’ equation is satisfi ed.

Nl 5 2d sin, (2.3)

where N is the order of refl ections (taken to be 1),  is the wavelength of the X-ray beam, d is the interplanar spacing, and  is the angle of incidence. Orders of refl ection greater than 1 are accounted for by the Miller indices.

The data generated by an X-ray diffraction experiment consist of a mea-surement of the intensity readings in the detector as a function of angle of incidence. The angle is generally reported as 2, since both the source and detector are at an angle . When there is no constructive interference, noth-ing but background scatter is detected. At 2 values at which constructive interference occurs, an increased radiation level is detected. A typical X-ray diffraction reading is shown in Figure 2-12.

Each peak in the diffractogram corresponds to a different plane in the crys-tal. Many calculations can be done using X-ray diffractograms and the Bragg

| Diffraction | The interaction of waves.

| Constructive Interference | The increase in amplitude resulting from two or more waves interacting in phase.

| Destructive Interference | A nullifi cation caused by two waves interacting out of phase.

| Braggs’ Equation | Formula that relates interplanar spacing in a lattice to constructive interference of diffracted X-rays. Named after the father and son (W. H. and

Figure 2-11 X-ray Diffraction Equipment Operation Figure 2-9 X-rays

Striking a Crystal Lattice

equation. If the wavelength of the X-ray source is known, then the Bragg equa-tion may be rearranged to determine the interplanar spacing of the plane cor-responding to each peak,

d5 nl

2 sin. (2.4)

Because many planes are present in a given crystal, they are identifi ed by their corresponding Miller indices (h k l). Thus Equation 2.3 could be written more properly as

d_hkl5 nl

2 sin. (2.5)

The interplanar spacing (d_hkl) of any given plane in a pure cubic system can be related to the lattice parameter by the following equation:

d_hkl5 a₀

"h²1 k²1 l². (2.6)

Regardless of the specifi c atoms, the existence of any specifi c plane is a func-tion of the lattice types. A specifi c set of h²1 k²1 l² combinations (called extinction conditions) exists that is the same for every simple cubic lattice;

a different set is present for every face-centered cubic lattice; and a third set exists for every body-centered cubic lattice. Table 2-5 summarizes the refl ec-tions present for each cubic lattice type.

Table 2-5 shows that the sum of the Miller indices 1h²1 k²1 l²2 for the plane that produces the fi rst diffraction peak in a BCC system must equal 2. Therefore, Equation 2.6 can be used to calculate the lattice parameter for each plane, if the lattice type is known. For any cubic lattice, the lattice parameter should be the same in each direction, which provides a check of other calculations.

The diffractogram also can be used to determine the type of lattice present in the material. If Equations 2.5 and 2.6 are combined, this relationship results:

sin² 5 l²

4a²₀1h²1 k²1 l²2. (2.7)

| Extinction Conditions | The systematic reduction in intensity of diffraction peaks from specifi c lattice planes.

Intensity

2u

Figure 2-12 Sample X-Ray Diffractogram

2.7 | X-Ray Diffraction 47

Table 2-5 Refl ections Present for Each Cubic Lattice Type

Lattice Type h²1 k²1 l²

BCC 2, 4, 6, 8, 10, 12, 14, 16

FCC 3, 4, 8, 11, 12, 16

Simple 1, 2, 3, 4, 5, 6, 8

Example 2-6

An X-ray diffraction source with a wavelength of 0.7107 angstroms is beamed through a sample to generate the following peaks. If the material has a BCC lattice structure, determine the interplanar spacing, the lattice parameter, and the sum of the squares of the Miller indices for each plane.

Peak 2

1 20.20

2 28.72

3 35.36

SOLUTION

For any given peak, the interplanar spacing is given by Equation 2.5, d_hkl5 nl/2 sin, so dhkl5 (1)(0.7107 angstroms)/2 sin (10.10°)  2.026 angstroms.

For a BCC system, Table 2-5 shows that the sum of the squares of the Miller indices for the fi rst plane should be 2. Thus, Equation 2.6 may be used to calculate the lattice parameter (a₀):

d_hkl5 a₀

"1h²1 k²1 l²2 12.026 angstroms2 5 a₀

!2

a₀5 2.868 angstroms to give the results summarized next:

Peak 2 1°2 d_hkl

1angstroms2 h²1 k²1 l² a₀ 1angstroms2

1 20.20 2.026 2 2.867

2 28.72 1.432 4 2.865

3 35.36 1.170 6 2.867

Equation 2.7 can be used to analyze the relationship between peaks. Because the X-ray wavelength does not change and the lattice parameter is the same for all planes in a cubic lattice, Equation 2.7 can be applied to two peaks to provide

sin²2

sin²1

5 1h²1 k²1 l²2₂ 1h²1 k²1 l²21

. (2.8)

The ratio of the sin² terms on the left side of Equation 2.8 gives the rela-tive ratio of the sums of the squares of the Miller indices of the two peaks.

Along with the information in Table 2-5, this information can identify the peak. Example 2-7 illustrates this more clearly.

Example 2-7

Determine the type of lattice in the material responsible for the following diffractogram information:

Peak 2

1 20.20

2 28.72

3 35.36

4 41.07

5 46.19

6 50.90

7 55.28

8 59.42

SOLUTION

We begin by applying Equation 2.8 to the fi rst two peaks. Note that the value listed in the table is 2, while  is needed for the equation.

sin²114.362 sin²110.102 5

1h²1 k²1 l²22

1h²1 k²1 l²21

, which provides

0.0615

0.03085 1h²1 k²1 l²21

1h²1 k²1 l²2₂5 2.

This tells us that the ratio of the sum of the squares of the Miller indi-ces of the fi rst two peaks is 2. There is no way to know the sum of the refl ections for the fi rst peak, but we have just determined that the 1h²1 k²1 l²2 for the second peak is twice that of the fi rst. According to Table 2-5, the 1h²1 k²1 l²2 values for the fi rst two peaks of a BCC system are 2 and 4. The ratio of ⁴₂ is 2, so the system could be a BCC system. Similarly, simple cubic has 1h²1 k²1 l²2 values of 1 and 2 for its fi rst two peaks, so simple cubic remains a possibility. However, the fi rst two FCC peaks have 3 and 4 as their values. This provides a ratio of 1.33 rather than 2, telling us that the diffractogram could not have been generated by an FCC lattice.

2.7 | X-Ray Diffraction 49

Until now, much of the discussion has treated materials as if they were comprised of a single, perfectly aligned crystal lattice. Instead, real materi-als consist of crystalline regions, or crystallites, separated from each other by grain boundaries. Therefore, a real material exhibits a structure much more like the crystal mosaic shown in Figure 2-13.

The same X-ray diffractogram provides information about the aver-age crystallite size. If the material was a pure crystal, each peak would be extremely thin and have virtually no spreading. In reality, each peak spreads across a range of 2 values. For relatively small grains, the amount of spreading is related to the thickness of crystallites in a plane by the Scherrer equation:

t5 0.9 B cos B

, (2.9)

where t represents the crystallite thickness,  is the wavelength of the X-ray source, B is the spread in the peak, and B is the  value at the top of the peak.

Because a diffraction peak narrows as it approaches the top, the spread in the peak depends on where it is measured. As a standard, full-width half-maximum (FWHM) is used, meaning that the spread in the peak is measured at the inten-sity value corresponding to half the highest value in the peak. Figure 2-14 shows FWHM for a sample peak.

By reading the values of 21 and 22 at FWHM, B can be determined from the equation

B5 0.51222 212 5 22 1 (2.10) Now compare each peak in the diffractogram to the fi rst peak.

Peak 2 Sin² Sin²/Sin²1 value that was eight times that of peak 1, according to Table 2-5.

Instead, peak 7 has an 1h²1 k²1 l²2 value that is seven times that of peak 1. The only pattern that would yield this exact ratio would be body-centered cubic, for which the 1h²1 k²1 l²2 is 2 for peak 1 and 14 for peak 7.

| Crystallites | Regions of a material in which

the atoms are arranged in a regular pattern.

| Grain Boundaries | The areas of a material

that separate different crystallite regions.

| Crystal Mosaic | A hypothetical structure accounting for irregularities in the boundaries between crystallites.

| Scherrer Equation | A means of relating the amount of spreading in a X-ray diffractogram to the thickness of the crystallites in the sample.

| Full-Width Half-Maximum

(FWHM) | A standard used to measure

the spread in the peak of a diffractogram, measured at the intensity value corresponding to the half highest value in the peak.

Intensity

21

2b

FWHM

22

2

2.7 | X-Ray Diffraction 51

Figure 2-13 Crystal Mosaic Structure

Figure 2-14 Full-Width Half-Maximum Measurement

| Optical Microscopy | The use of light to magnify

objects up to 2000 times.

Example 2-8

Estimate the thickness of the crystallites from the planes corresponding to peak 2 in Example 2-6 given that 215 28.46 and 225 28.98.

SOLUTION

Crystallite thickness is estimated using the Scherrer equation (2.9), so t5 0.9l

B cosB

t5 0.91.7107 nm2

0.5128.98 2 28.462 cos 128.72/22 t5 2.54 nm

2.8 MICROSCOPY

S

eeing the features directly often would be the best way to understand the structure of a material. Some features, including grain sizes, large fl aws in the material, cracking, and structures present in alloys, at times are visible to the naked eye. Often, however, the features of interest are too small to be seen directly. In such cases, the use of microscopes becomes valuable.

Several different types of microscopes exist and are classifi ed by their source of light (or other radiation). The most common microscopes (present in essen-tially every science lab) are optical microscopes. If the material is opaque (e.g., metals, ceramics, and most polymers and composites), only a surface can be examined microscopically, and refl ected light passing through the lens must reveal the image. For most materials, the surface must be polished before any meaningful features will be revealed. Many materials require surface treat-ment by an etching agent to reveal information. The reactivity between etching agents and some materials varies depending on the orientation of their grains.

Specifi c etching agents are chosen so that adjoining grains will be affected dif-ferently and the contrast between the grains will become visible under the opti-cal microscope.

Optical microscopy offers several advantages. The equipment is inex-pensive and easy to operate. Large features, such as grains and cracks, are often apparent. Commercial software can calculate the size of each visible grain. However, optical microscopes are limited to about 20003 magnifi ca-tion, and many of the features that govern behavior exist at a much smaller scale.

When optical microscopy is insuffi cient, materials scientists turn to elec-tron microscopy. Here, instead of visible light, a focused high-energy beam of electrons serves as the source for the image. The effective wavelength of an electron beam is 0.003 nm, allowing for resolution of fi ner details.

Two distinct types of electron microscopes are available to provide differ-ent information.

Scanning electron microscopes (SEMs) collect the back-scattered beam of electrons and use it to project a blown-up image on a monitor, much like a television or computer screen. Resolution of details to the submicron level is possible with an SEM, and most systems are capable of capturing the digital image for printing or analysis. Surface features are directly visible, which makes the SEM ideal for rough surfaces, even at lower magnifi cation. When using the SEM, some skill in positioning and focusing the beam is required. Moreover, researchers must be careful to ensure that the portion of the material they are examining is suffi ciently representative of the entire material. Unlike an optical microscope, a high-end SEM costs several hundred thousand dollars.

Transmission electron microscopy (TEM) involves passing the electron beam through the sample and using the differences in beam scattering and dif-fraction to resolve an image. TEM is especially effective for examining micro-structural defects. TEMs can magnify an image 1,000,000 times, but, like SEMs, they are extremely expensive and require some skill to operate. Addi-tionally, sample preparation presents challenges since most materials absorb electron beams. To compensate, an extremely thin fi lm of material must be pre-pared for examination to allow enough of the electron beam to pass through.

Because the TEM uses diffraction of an electron beam to gather its imaging information, examination of the diffraction and scattering patterns can pro-vide additional structural information. The principles of electron diffraction are highly similar to those of X-ray diffraction, discussed earlier in this chapter.

In document PONTIFICIA UNIVERSIDAD CATÓLICA DEL PERÚ FACULTAD DE GESTIÓN Y ALTA DIRECCIÓN (página 99-104)